wrote in message
ups.com...
In air-filled rectangular waveguide, what is the meaning by " the
operating frequency is to be at least 25% greater than the cutoff
frequency of the dominant mode"?
Can show it in mathematically?
Thanks

Laura


As frequency decreases wavelength increases, Cutoff is the longest
wavelength that will fit in the pipe.
Fcutoff = speed of light/2 times the largest crosssectional dimension of the
waveguide

or Fc= C/2d

However the wavlength of a frequency in free space is significanty
shorter than the wavelength inside the waveguide so the actual Fc will
be greater than that calculated usng freespace values. My slide rule says
that if the angle of incidence inside the waveguide is 45 degrees and the
free space wavelenghth is .05 meters wavelength inside the waveguide is
about .075 meters. Dont hold me to these numbers but they should be
reasonably close. Im not very good at writting equations using a computer
but the Practical Antenna Handbook by Joeseph Carr has a pretty good
discription of whats going on. I am sure others will give you more
references and probably more precise numbers.

Jimmie